I'm always trying to find the integral representation of $\pi$ using some interesting special function, at this time I have got the below representation
$$I=\int_{-2}^{2} \tan^{-1} \bigg( \exp(-x^2\text{erf}(x)) \bigg) \;dx=\pi$$ and according to Wolfram alpha its numerical value is very close to $\pi$.
The problem that I have accrossed is the closed form of : $\exp(-x^2\text{erf}(x))$ in the range $[-2,2]$ , I have used $$\mathrm{erf}\!\left(x\right)^2\approx1-\exp\Big(-\frac 4 {\pi}\,\frac{1+\alpha x^2}{1+\beta x^2}\,x^2 \Big)$$ $$\alpha=\frac{10-\pi ^2}{5 (\pi -3) \pi }$$ $$\beta=\frac{120-60 \pi +7 \pi ^2}{15 (\pi -3) \pi }.$$ The value of the corresponding error function is $1.1568\times 10^{-7}$ that is to say almost $250$ times smaller than with the initial formulation. The maximum error is $0.00035$. But when we try to replace that approximation in $\tan^{-1}$ for evaluation the formula would be complicat to integrate it. My question here is if this result is known. Is it $\pi$? If no any simple way for integration ?
Best Answer
Yes, the integral $I$ is equal to $\pi$. Note that for $t>0$ $$\arctan(t)+\arctan(1/t)=\pi/2.$$ and after letting $y=-x$ we get $$I:=\int_{-2}^{2} \tan^{-1} \bigg( \exp(-x^2\text{erf}(x)) \bigg) \;dx =\int_{-2}^{2} \tan^{-1} \bigg( \exp(y^2\text{erf}(y)) \bigg) \;dy.$$ Hence $$I=\frac{1}{2}\int_{-2}^{2}\arctan(t(x))+\arctan(1/t(x))dx=\frac{\pi/2\cdot 4}{2}=\pi$$ where $t(x)=\exp(-x^2\text{erf}(x))$.
The same argument holds if we replace $-x^2\text{erf}(x)$ with any odd function (see Michael Seifert's comment below).